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OLEGan [10]
3 years ago
5

Systems w/ substitution y=4x+6 & y=-5+21

Mathematics
1 answer:
Law Incorporation [45]3 years ago
7 0
So basically what you would is u would plug the first equation into the second equation. -5x+21=4x+6
                      -21      -21
-5x=4x-15
-4x   -4x
-9x=-15
-9x/-9=15/-9
x=-1.66
then you would plug x back into the equation to get your y value
y=4(-1.66)+6
y=6.64+6
y=12.64



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Write the expression using rational exponents. Then simplify and convert back to radical notation.
ioda

Answer:

The radical notation is 3x\sqrt[3]{y^2z}

Step-by-step explanation:

Given

\sqrt[3]{27 x^{3} y^{2} z}

Step 1 of 1

Write the expression using rational exponents.

\sqrt[n]{a^{m}}=\left(a^{m}\right)^{\frac{1}{n}}

=a^{\frac{m}{n}}:\left({27 x^{3} y^{2} z})^{\frac{1}{3}}

$(a \cdot b)^{r}=a^{r} \cdot b^{r}:(27)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$

=$(3^3)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$

$=\left(3\right)\left(x}\right)} \cdot\left(y}\right)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

Simplify $3 x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3 x \sqrt[3]{y^{2} z}$

Learn more about radical notation, refer :

brainly.com/question/15678734

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3 years ago
What is the distance between points v (3,3) and w (-2,-3) round to the nearest tenth if necessary
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Answer:

7.81 u

Step-by-step explanation:

<u>Given :- </u>

  • Two points (3,3) and (-2,-3) is given to us.

And we need to find out the distance between the two points . So , here we can use the distance formula to find out the distance. As,

:\implies D = √{(x2-x1)² + (y2-y1)²}

:\implies D =√[ (3+2)² +(-3-3)²]

:\implies D =√[ 5² +6²]

:\implies D =√[ 25 +36]

:\implies D = 61

:\implies D = 7.81

<h3>Hence the distance between the two points is 7.81 units .</h3>
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m_a_m_a [10]

Answer:

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if we use both conditions, we will have:

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Hope this helps.

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